Question: $\text E = \left[\begin{array}{rrr}1 & -1 & 5 \\ 5 & 5 & 0\end{array}\right]$ and $\text F = \left[\begin{array}{rr}-2 & -1 \\ 5 & 2 \\ 5 & -2\end{array}\right]$ Let $\text {H = EF}$. Find $\text H$. $ {H = }$
Explanation: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{E}$ and the first column of $\text{F}$. $ \text {H}=\left[\begin{array}{rr}{1} & {-1} & {5} \\ 5 & 5 & 0\end{array}\right]\left[\begin{array}{rr} {-2} & -1 \\ {5} & 2 \\ {5} & -2\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(1,-1,5)\cdot(-2,5,5)\\\\ &=1 \cdot -2-1\cdot 5 + 5\cdot 5\\\\ &=18 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $5 \cdot -2+5\cdot 5 + 0\cdot 5 = 15$ (Choice B) B $-1 \cdot 2 +5\cdot -1 = -7$ (Choice C) C $1 \cdot -1-1\cdot 2 + 5\cdot -2 = -13$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}=\left[\begin{array}{rr}18 & -13 \\ 15 & 5\end{array}\right]$